8

fine the limit :

$$\lim_{x \to 0}\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\rfloor=?$$

We denote the floor funtion by $\lfloor x\rfloor$. My try:

\begin{align} \lim_{x \to 0}\frac{\tan^{n}x - \sin^{n} x}{x^{n + 2}} &= \lim_{x \to 0}\frac{\tan x - \sin x}{x^{3}}\cdot \sum_{i = 0}^{n - 1}\frac{\tan^{n - 1 - i}x}{x^{n - 1 - i}}\cdot\frac{\sin ^{i}x}{x^{i}}\\ &= \frac{1}{2}\cdot\sum_{i = 0}^{n - 1}1\\ &= \frac{n}{2} \end{align} So: \begin{align} \lim_{x \to 0}\frac{\tan^{98}x - \sin^{98} x}{x^{100}}&=49\\ \lim_{x \to 0}\left\lfloor\frac{\tan^{98}x - \sin^{98} x}{x^{100}}\right\rfloor&=49 \end{align}

is this correct?

user402750
  • 173
  • I don't really understand the step to the second line, but wolfram alpha agrees with you so chances are you're correct. Perhaps elaborating on that step won't make the proof more correct, but certainly more readable and understandable. –  Mar 08 '17 at 10:22
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    Yes it is correct. @vrutehagel: he used the identity $a^n-b^n=(a-b)\cdot\sum_{i=0}^{n-1}a^{n-1-i}b^i$. – Maczinga Mar 08 '17 at 10:41
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    But if $\frac{\tan^{98}x - \sin^{98} x}{x^{100}}$ is always less than 49 when $x$ is near 0, then the value of limit will be $48$. You need to check it. – guest Mar 08 '17 at 10:43

2 Answers2

6

Using the Taylor series $$ \begin{aligned} \tan x&=x+\frac13x^3+\frac2{15}x^5+\cdots,\\ \sin x&=x-\frac16x^3+\frac1{120}x^5+\cdots \end{aligned} $$ and the binomial formula, we arrive at $$ \tan^{98}x=x^{98}+\frac{98}3x^{100}+[\frac19\cdot\binom{98}2+\frac{98\cdot2}{15}]x^{102}+\cdots $$ and $$ \sin^{98}x=x^{98}-\frac{98}6x^{100}+[\frac1{36}\cdot\binom{98}2+\frac{98}{120}]x^{102}+\cdots. $$ Therefore $$ \tan^{98}x-\sin^{98}x=49x^{100}+\frac{1225}3x^{102}+\cdots, $$ and $$ \frac{\tan^{98}x-\sin^{98}x}{x^{100}}=49+\frac{1225}3x^2+\cdots. $$ When $x$ is close enough to zero, that quadratic term dominates the (possibly negative) cut-off terms. Therefore, when $|x|$ is small enough we have $$ 49\le \frac{\tan^{98}x-\sin^{98}x}{x^{100}}<50. $$ The answer is thus $49$.

Jyrki Lahtonen
  • 133,153
2

No, your last transformation (taking the floor in and out of the limit) is not justified.

As a counterexample, consider

$$\lim_{n\to\infty}\left(49-\frac1n\right)=49$$

versus

$$\lim_{n\to\infty}\left\lfloor49-\frac1n\right\rfloor=48.$$